Convergence of $a_1=1,a_2=\sqrt{7},a_3=\sqrt{7\sqrt{7}}, a_4=\sqrt{7\sqrt{7\sqrt{7}}}, \dots$ Define a sequence $a_n$ as follows: $a_1=1,a_2=\sqrt{7},a_3=\sqrt{7\sqrt{7}}, a_4=\sqrt{7\sqrt{7\sqrt{7}}}, \dots$
Determine if it's convergent and find its limit.
The sequence satisfies $a_n=\sqrt{7a_{n-1}}$. If it's convergent with limit $a$, then $a^2=7a$, so either $a=0$ or $a=7$. We show that the sequence is monotonically increasing and bounded above by $7$ so its limit cannot be equal $0$, thus it is $7$.
Claim. $a_n < 7$ for all $n$. 
Proof: Induction on $n$. The base is clear. Assume $a_{n-1} < 7$. This is equivalent to saying that $7a_{n-1} < 49$, which happens iff $\sqrt {7a_{n-1}} < 7$. Then $a_n=\sqrt{7a_{n-1}} < 7$.
It remains to show $a_n$ is monotonically increasing.
Consider 
$a_n/a_{n-1}=\sqrt{7/a_{n-1}}$. We prove that this is greater than $1$. It will follow that $a_n > a_{n-1}$.
Since $a_n < 7$, $7/a_{n-1}> 1$, whence $a_n/a_{n-1} > 1$. Thus the sequence is increasing and bounded above, so it's convergent. It's limit cannot be zero, so it must be $7$.
Is this a correct proof?
 A: It all boils down to showing that $$0<a_n<7\implies a_n<\sqrt{7a_n}=a_{n+1}<7,$$ which is fairly obvious (geometric average).
A: Here's a totally different approach.
Let $T(a)=\sqrt{7a}$. For $a \geq 2$,
$$
T(a) \geq \sqrt{14} > \sqrt{4} = 2
\qquad
\text{and}
\qquad
|T^{\prime}(a)| = \frac{\sqrt{7}}{2\sqrt{a}} \leq \frac{\sqrt{7}}{2\sqrt{2}} = \frac{\sqrt{14}}{4} < \frac{\sqrt{16}}{4} = 1.
$$
Therefore, $T$ is a contraction on $[2, \infty)$.
By the Banach fixed point theorem, $T$ has a fixed point in $X$.
You already proved that this fixed point has to be $a=7$ (since $a=0 \notin X$).
Conclude by noting that $T(a_1)=T(1)\geq2$.
A: Technically you have merged couple of theorems and only the first one should be sufficient. Even though @parsiad is saying that his approach is different, that one is included in your proof. If you are to use fixed point then you have to prove that you are dealing with contraction within a complete metric space as that is what Banach fixed point theorem is saying. So all you need is to define $T(x)=\sqrt{7x}$ over for example $x > \frac{5}{2}$. Then prove this is a contraction and you are there. In that case you do not need other two steps.
If you want to go another way which is using monotonic theorem then you've proved correctly that it is monotonically increasing and that it is bounded. All you need to prove is that it can get as close as you want to $7$.
However, in your proof you have mixed the two. You have taken one part of one theorem for which you need to prove contraction and you did not, and you have taken another theorem for which you need to prove that $7$ is a limit and you did not.
Essentially the fact that you can calculate the fixed point is only suggesting that this might be a fixed point. The fact that you can calculate a fixed point does not suggest that this is indeed the fixed point. For example you might calculate a fixed point assuming it is real while your sequence slips into the complex realm.
So your proof is not sufficiently precise.
Imagine you have a situation where
$$a_1=1,a2=\sqrt{6|7},a3=\sqrt{6|7\sqrt{6|7}}...$$
$6|7$ is taken at random to be either $6$ or $7$
You could prove that this is a contraction for any series that is consisting of $6$ or $7$ or both. This means that there is a fixed point, but you cannot find it by solving any equation per se.
If you take the approach for monotonically increasing series that one is working as well, yet again you cannot find the limit itself.
In both cases you are only sure it exists.
For this last the convergence reaches $7\left ( \frac{6}{7} \right )^x$ where $0 \leq x \leq 1$ depending on the distribution of $6$'s and $7$'s.
If you, however, try to find a fixed point by solving the "equation", this is what you get:
$$7\left ( \frac{6}{7} \right )^x=\sqrt{6|7\cdot7\left ( \frac{6}{7} \right )^x}$$
$$7^2\left ( \frac{6}{7} \right )^{2x}=6|7\cdot7\left ( \frac{6}{7} \right )^x$$
$$\left ( \frac{6}{7} \right )^{x}=\frac{6}{7}|1 \; \; \; \text{what?!}$$
(This last has some mathematical sense only if $x=0$ or $x=1$ yet from our definition you cannot assume either is true.)
In order to have the fixed point as a solution of the equation, you need already all the assumptions about having the fixed point.
